Part IV: Complete Solutions, Chapter 9
486
Chapter 9: Hypothesis Testing
Section 9.1
Note: For all graphs provided, the P value is indicated by the shaded portion in the tails.
1.
See text for definitions. Essays may include
(a) A working hypothesis about the population parameter in question is called the null hypothesis. The
value specified in the null hypothesis is often a historical value, a claim, or a production specification.
(b) Any hypothesis that differs from the null hypothesis is called an alternate hypothesis.
(c) If we reject the null hypothesis when it is in fact true, we have an error that is called a type I error. On
the other hand, if we fail to reject the null hypothesis when it is in fact false, we have made an error
that is called a type II error.
(d) The probability with which we are willing to risk a type I error is called the level of significance of a
test. The probability of making a type II error is denoted by  .
2.
The alternate hypothesis is used to determine which type of test is used.
3.
No, if we fail to reject the null hypothesis, we have not proven it to be true beyond all doubt. The evidence
is not sufficient to merit rejecting H 0 .
4.
No, if we reject the null hypothesis, we have not proven it to be false beyond all doubt. The test was
conducted with a level of significance  that is the probability with which we are willing to risk a type I
error (rejecting H 0 when it is in fact true).
5.
(a) The claim is  = 60 kg, so you would use H 0 :  = 60 kg.
(b) We want to know if the average weight is less than 60 kg, so you would use H1:  < 60 kg.
(c) We want to know if the average weight is greater than 60 kg, so you would use H1:  > 60 kg.
(d) We want to know if the average weight is different from 60 kg, so you would use H1:   60 kg.
(e) Since part (b) is a left-tailed test, the critical region is on the left. Since part (c) is a right-tailed test, the
critical region is on the right. Since part (d) is a two-tailed test, the critical region is on both sides of
the mean.
6.
(a) The claim is  = 8.3 min, so you would use H 0 :  = 8.3 min. If you believe that the average is less than
8.3 min, then you would use H1:  < 8.3 min. This is a left-tailed test.
(b) The claim is  = 8.3 min, so you would use H 0 :  = 8.3 min. If you believe the average is different
from 8.3 min, then you would use H1:   8.3 min. This is a two-tailed test.
(c) The claim is  = 4.5 min, so you would use H 0 :  = 4.5 min. If you believe the average is more than
4.5 min, then you would use H1:  > 4.5 min. This is a right-tailed test.
(d) The claim is  = 4.5 min, so you would use H 0 :  = 4.5 min. If you believe the average is different
from 4.5 min, then you would use H1:   4.5 min. This is a two-tailed test.
7.
(a) The claim is  = 16.4 ft, so H 0 :  = 16.4 ft.
(b) You want to know if the average is getting larger, so H1:  > 16.4 ft.
(c) You want to know if the average is getting smaller, so H1:  < 16.4 ft.
(d) You want to know if the average is different from 16.4 ft, so H1:   16.4 ft.
(e) Since part (b) is a right-tailed test, the area corresponding to the P value is on the right. Since part (c) is
a left-tailed test, the area corresponding to the P value is on the left. Since part (d) is a two-tailed test,
the area corresponding to the P value is on both sides of the mean.
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Part IV: Complete Solutions, Chapter 9
8.
487
(a) The claim is  = 8.7 s, so H 0 :  = 8.7 s.
(b) You want to know if the average is longer, so H1:  > 8.7 s.
(c) You want to know if the average is reduced, so H1:  < 8.7 s.
(d) Since part (b) is a right-tailed test, the P value area is on the right. Since part (c) is a left-tailed test, the
P value area is on the left.
9.
(a)   0.01
H 0 :   4.7%
H1:   4.7%
Since > is in H1 , use a right-tailed test.
(b) Use the standard normal distribution. We assume x has a normal distribution with known standard
deviation  . Note that  is given in the null hypothesis.
z
x 


n
5.38  4.7
2.4
10
 0.90
(c) P value = P(z > 0.90) = 0.1841
(d) Since a P value of 0.1841 > 0.01 for , we fail to reject H 0 . The data are not statistically significant.
(e) There is insufficient evidence at the 0.01 level to reject the claim that average yield for bank stocks
equals average yield for all stocks.
10. (a)   0.05
H 0 :   85 mg/100 mL
H1:   85 mg/100 mL
Since > is in H1 , use a right-tailed test.
(b) Use the standard normal distribution. We assume that x has a normal distribution with known standard
deviation  . Note that  is given in the null hypothesis.
z
x 

n

93.8  85
12.5
8
 1.99
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(c) P value = P(z > 1.99) = 0.0233
(d) Since a P value of 0.0233  0.05, reject H 0 . Yes, the data are statistically significant.
(e) The sample evidence is sufficient at the 0.05 level to justify rejecting H 0 . It seems Gentle Ben’s
glucose is higher than average.
11. (a)   0.01
H 0 :   4.55 g
H1:   4.55 g
Since < is in H1 , use a left-tailed test.
(b) Use the standard normal distribution. We assume that x has a normal distribution with known standard
deviation  . Note that  is given in the null hypothesis.
z
x 

n

3.75  4.55
0.70
6
 2.80
(c) P value = P(z < 2.80) = 0.0026
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(d) Since a P value of 0.0026  0.01, we reject H 0 . Yes, the data are statistically significant.
(e) The sample evidence is sufficient at the 0.01 level to justify rejecting H 0 . It seems the humming birds
in the Grand Canyon region have a lower average weight.
12. (a)   0.05
H 0 :   19
H1:   19
Since < is in H1 , use a left-tailed test.
(b) Use the standard normal distribution. We assume that x has a normal distribution with known standard
deviation  . Note that  is given in the null hypothesis.
z
x 


n
17.1  19
4.5
14
 1.58
(c) P value = P(z < 1.58) = 0.0571
(d) Since a P value of 0.0571 > 0.05 for , we fail to reject H 0 . The data are not statistically significant.
(e) There is insufficient evidence at the 0.05 level to reject H 0 . It seems the average P/E for large banks is
not less than that of the S&P 500 Index.
13. (a)   0.01
H 0 :   11%
H1:   11%
Since  is in H1 , use a two-tailed test.
(b) Use the standard normal distribution. We assume that x has a normal distribution with known standard
deviation  . Note that  is given in the null hypothesis.
z
x 

n

12.5  11
5.0
16
 1.20
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(c) P value = 2P(z > 1.20) = 2(0.1151) = 0.2302
(d) Since a P value of 0.2302 > 0.01 for , we fail to reject H 0 . The data are not statistically significant.
(e) There is insufficient evidence at the 0.01 level of significance to reject H 0 . It seems the average hail
damage to wheat crops in Weld County matches the national average.
14. (a)   0.01
H 0 :   28 mL/kg
H1:   28 mL/kg
Since  is in H1 , use a two-tailed test.
(b) Use the standard normal distribution. We assume that x has a normal distribution with known standard
deviation  . Note that  is given in the null hypothesis.
z
x 

n

32.7  28
4.75
7
 2.62
(c) P value = 2P(z > 2.62) = 2(0.0044) = 0.0088
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(d) Since a P value of 0.0088  0.01, we reject H 0 . The data are statistically significant.
(e) At the 1% level of significance, the sample average is sufficiently different from  = 28 that we reject
H 0 . It seems Roger’s average red cell volume is different from the average for healthy adults.
Section 9.2
1.
The P value for a two-tailed test of μ is twice the P value for a one-tailed test based on the same sample
data and null hypothesis.
2.
If σ is known, use the standard normal distribution. If σ is not known, use the Student’s t distribution with n
– 1 degrees of freedom.
3.
Use n – 1 degrees of freedom.
4.
Not necessarily. If the P value = 0.04, you would reject at the α = 0.05 level but not at the α = 0.01 level.
On the other hand, if the P value is 0.003, you would reject at both the α = 0.05 and α = 0.01 levels.
5.
Yes. If the P value is less than α = 0.01, then it also will be less than α = 0.05. In both cases, reject the null
hypothesis.
6.
No. The P value for the two-tailed test is twice the P value for the one-tailed test. If the P value < 0.01, then
2 × P value might be greater than or less than 0.01.
7.
(a)   0.01
H 0 :   16.4 ft
H1:   16.4 ft
(b) Use the standard normal distribution. The sample size is large, n  30, and  = 3.5.
x   17.3  16.4
z

 1.54

n
3.5
36
(c) P value = P(z > 1.54) = 0.0618
(d) Since a P value of 0.0618 > 0.01, we fail to reject H 0 . The data are not statistically significant.
(e) At the 1% level of significance, there is insufficient evidence to say average storm level is increasing.
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8.
(a)   0.01
H 0 :   38 h
H1:   38 h
(b) Use the standard normal distribution. The sample size is large, n  30, and σ = 1.2 hours.
x   37.5  38
z

 2.86

n
1.2
47
(c) P value = P(z < 2.86) = 0.0021
(d) Since a P value of 0.0021  0.01, we reject H 0 . The data are statistically significant.
(e) At the 1% level of significance, evidence shows average assembly time is less than 38 hours.
9.
(a)   0.05
H 0 :   41.7
H1:   41.7
(b) Use a standard normal distribution. The sample size is large, n  30, and σ = 18.5 e-mails.
x   36.2  41.7
z

 1.99

n
18.5
45
(c) P value = 2P(z < 1.99) = 2(0.0233) = 0.0466
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(d) Since a P value of 0.0466  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, there is sufficient evidence to say the average number of e-mails is
different than 41.7 under the new priority system.
10. (a)   0.05
H 0 :   7.4 pH
H1:   7.4 pH
(b) Use the Student’s t distribution with d.f. = n  1 = 31  1 = 30. The sample size is large, n  30, and 
is unknown.
x   8.1  7.4
t

 2.051
s
n
1.9
31
(c) If d.f. = 30, 2.051 falls between the entries 2.042 and 2.457, use two-tailed areas to find that
0.020 < P value < 0.050. Using a TI-84, P value ≈ 0.0491.
(d) Since the entire P-value interval  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence is sufficient to say that the drug has changed the mean pH
level.
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11. (a)   0.01
H 0 :   1.75 years
H1:   1.75 years
(b) Use the Student’s t distribution with d.f. = n  1 = 46  1 = 45. The sample size is large, n  30, and 
is unknown.
x   2.05  1.75
t

 2.481
s
n
0.82
46
(c) For d.f. = 45, 2.481 falls between entries 2.412 and 2.690. Use one-tailed areas to find that
0.005 < P value < 0.010. Using a TI-84, P value ≈ 0.0084.
(d) Since the entire P-value interval  0.01, we reject H 0 . Yes, the data are statistically significant.
(e) At the 1% level of significance, sample data indicate that the average age of Minnesota region coyotes
is greater than 1.75 years.
12. (a)   0.05
H 0 :   19 in.
H1:   19 in.
(b) Use the Student’s t distribution with d.f. = n  1 = 51  1 = 50. The sample size is large, n  30, and 
is unknown.
x   18.5  19
t

 1.116
s
n
3.2
51
(c) For d.f. = 50, 1.116 falls between entries 0.679 and 1.164.
Use one-tail areas to find that 0.125 < P value < 0.250. Using a TI-84, P value ≈ 0.1349.
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(d) Since the P-value interval > 0.05, we fail to reject H 0 . The data are not statistically significant.
(e) At the 5% level of significance, the sample data do not indicate that the average fish length is less than
19 inches.
13. (a)   0.05
H 0 :   19.4
H1:   19.4
(b) Use the Student’s t distribution with d.f. = n  1 = 36  1 = 35. The sample size is large, n  30, and 
is unknown.
x   17.9  19.4
t

 1.731
s
n
5.2
36
(c) For d.f. = 35, 1.731 falls between entries 1.690 and 2.030. Use two-tailed areas to find that
0.05 < P value < 0.100. Using a TI-84, P value ≈ 0.0923.
(d) Since the P-value interval > 0.05, we fail to reject H 0 . The data are not statistically significant.
(e) At the 5% level of significance, the sample evidence does not support rejecting the claim that the
average P/E for socially responsible funds is different from that of the S&P stock index.
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14. (a)   0.01
H 0 :   8.0 ppb
H1:   8.0 ppb
(b) Use the Student’s t distribution with d.f. = n  1 = 37  1 = 36. The sample size is large, n  30, and 
is unknown.
x   7.2  8.0
t

 2.561
s
n
1.9
37
(c) In Table 6, use the closest d.f. smaller than 36 or d.f. = 35. Since 2.561 falls between entries 2.438 and
2.724, use one-tailed areas to find 0.005 < P value < 0.010. Using a TI-84, P value ≈ 0.0074.
(d) Since the entire P-value interval  0.01, we reject H 0 . Yes, the data are statistically significant.
(e) At the 1% level of significance, sample data support the claim that the average arsenic content is less
than 8.0 ppb.
15. (i) Use a calculator to verify. Rounded values are used in part (ii).
(ii) (a)   0.05
H 0 :   4.8
H1:   4.8
(b) Use the Student’s t distribution with d.f. = n  1 = 6  1 = 5. We assume that x has a distribution
that is approximately normal and that  is unknown.
x   4.40  4.8
t

 3.499
s
n
0.28
6
(c) For d.f. = 5, 3.499 falls between entries 3.365 and 4.032. Use one-tailed areas to find that
0.005 < P value < 0.010. Using a TI-84, P value ≈ 0.0086.
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Student's t Distribution with d.f. = 5
-3.499
-2.000
-1.000
0.000
1.000
2.000
3.500
(d) Since the entire P-value interval  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, sample evidence supports the claim that the average RBC count
for this patient is less than 4.8.
16. (i) Use a calculator to verify. Rounded values are used in part (ii).
(ii) (a)   0.01
H 0 :   14
H1:   14
(b) Use the Student’s t distribution with d.f. = n  1 = 10  1 = 9. We assume that x has a distribution
that is approximately normal and that  is unknown.
x   15.1  14
t

 1.386
s
n
2.51
10
(c) For d.f. = 9, 1.386 falls between entries 1.383 and 1.574. Use one-tailed areas to find that
0.075 < P value < 0.100. Using a TI-84, P value ≈ 0.0996.
(d) Since P-value interval > 0.01, we fail to reject H 0 . No, the data are not statistically significant.
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(e) At the 1% level of significance, the sample data do not support the claim that the average HC for
this patient is higher than 14.
17. (i) Use a calculator to verify. Rounded values are used in part (ii).
(ii) (a)   0.01
H 0 :   67
H1:   67
(b) Use the Student’s t distribution with d.f. = n  1 = 16  1 = 15. We assume that x has a distribution
that is approximately normal and that  is unknown.
x   61.8  67
t

 1.962
s
n
10.6
16
(c) For d.f. = 15, 1.962 falls between entries 1.753 and 2.131. Use two-tailed areas to find that
0.050 < P value < 0.100. Using a TI-84, P value ≈ 0.0686.
(d) Since P-value interval > 0.01, we fail to reject H 0 . The data are not statistically significant.
(e) At the 1% level of significance, sample evidence does not support a claim that the average
thickness of slab avalanches in Vail is different from that of those in Canada.
18. (i) Use a calculator to verify. Rounded values are used in part (ii).
(ii) (a)   0.05
H 0 :   77 years
H1:   77 years
(b) Use the Student’s t distribution with d.f. = n  1 = 20  1 = 19. We assume that x has a distribution
that is approximately normal and that  is unknown.
x   71.4  77
t

 1.213
s
n
20.65
20
(c) For d.f. = 19, 1.213 falls between entries 1.187 and 1.328. Use one-tailed areas to find that
0.100 < P value < 0.125. Using a TI-84, P value ≈ 0.1200.
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(d) Since the P-value interval > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 5% level of significance, evidence is not strong enough to conclude that the population mean
life span is less than 77 years.
19. (i) Use a calculator to verify. Rounded values are used in part (ii).
(ii) (a)   0.05
H 0 :   8.8
H1:   8.8
(b) Use the Student’s t distribution with d.f. = n  1 = 14  1 = 13. We assume that x has a distribution
that is approximately normal and that  is unknown.
x   7.36  8.8
t

 1.337
s
n
4.03
14
(c) For d.f. = 13, 1.337 falls between entries 1.204 and 1.350. Use two-tailed areas to find that
0.200 < P value < 0.250. Using a TI-84, P value ≈ 0.2042.
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(d) Since P-value interval > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 5% level of significance, we cannot conclude that the catch is different from 8.8 fish per day.
20. (i) Use a calculator to verify. Rounded values are used in part (ii).
(ii) (a)   0.01
H 0 :   1,300
H1:   1,300
(b) Use the Student’s t distribution with d.f. = n  1 = 10  1 = 9. We assume that x has a distribution
that is approximately normal and that  is unknown.
x   1268  1300
t

 2.714
s
n
37.29
10
(c) For d.f. = 9, 2.714 falls between entries 2.262 and 2.821. Use two-tailed areas to find
0.020 < P value < 0.050. Using a TI-84, P value ≈ 0.0239.
(d) Since P-value interval > 0.01, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 1% level of significance, there is not enough evidence to conclude that the population mean
of tree ring dates is different from 1,300.
21. (a) The P value of a one-tailed test is smaller. For a two-tailed test, the P value is double because it
includes the area in both tails.
(b) Yes; the P value of a one-tailed test is smaller, so it might be smaller than , whereas the P value of a
two-tailed test is larger than .
(c) Yes; if the two-tailed P value is less than , the one-tail area is also less than .
(d) Yes, the conclusions can be different. The conclusion based on the two-tailed test is more conservative
in the sense that the sample data must be more extreme (differ more from H 0 ) in order to reject H 0 .
22. Essay or class discussion.
23. (a) H 0 :   20
H1:   20
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For  = 0.01, c = 1  0.01 = 0.99,   4, and zc  2.58.
E  zc

n
 2.58
4
36
 1.72
xExE
22  1.72    22  1.72
20.28    23.72
The hypothesized mean  = 20 is not in the interval. Therefore, we reject H 0 .
(b) Because n = 36 is large, the sampling distribution of x is approximately normal by the central limit
theorem, and we know σ.
x   22  20
z

 3.00
 / n 4/ 36
From Table 5, P value = 2P(z < 3.00) = 2(0.0013) = 0.0026. Since 0.0026  0.01, we reject H 0 . The
results are the same.
24. (a) H 0 :   21
H1:   21
For  = 0.01, c = 1  0.01 = 0.99,   4, and zc  2.58.
E  zc

n
 2.58
4
36
 1.72
xExE
22  1.72    22  1.72
20.28    23.72
The hypothesized mean  = 21 falls into the confidence interval. Therefore, we do not reject H 0 .
(b) For  = 0.01, the two-tailed test’s critical values are z0  2.58. Because n = 36 is large, the sampling
distribution of x is approximately normal by the central limit theorem, and we know σ.
x   22  21
z

 1.50
 / n 4/ 36
From Table 5, P value = 2P(z < 1.50) = 2(0.0668) = 0.1336. Since 0.1336 > 0.01, we do not
reject H 0 . The results are the same.
25. For a right-tailed test and  = 0.01, the critical value is z0  2.33; critical region is values to the right of
2.33. Since the sample statistic z = 1.54 is not in the critical region, fail to reject H 0 . At the 1% level, there
is insufficient evidence to say that the average storm level is increasing. Conclusion is the same as with the
P-value method.
26. For a left-tailed test and  = 0.01, the critical value z0  2.33; critical region is values to the left of 2.33.
Since the sample statistic z = 2.86 is in the critical region, reject H 0 . At the 1% level, evidence shows that
average assembly time is less than 38 hours. The conclusion is that same as with the P-value method.
27. For a two-tailed test and  = 0.05, the critical value z0  1.96; critical regions are values to the left of
1.96 and values to the right of 1.96. Since the sample test statistic z = 1.99 is in the critical region, reject
H 0 . At the 5% level, there is sufficient evidence to say the average number of e-mails is different with the
new priority system. The conclusion is the same as with the P-value method.
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Part IV: Complete Solutions, Chapter 9
502
28. For a two-tailed test and  = 0.05, critical values are t0  2.042 with d.f. = 30; critical regions are values
to the right of 2.042 and those to the left of 2.042. Since the sample test statistic t = 0.051 is in the critical
region, reject H 0 . At the 5% level, the evidence is sufficient to say that the drug has changed the mean pH
level. The conclusion is the the same as with the P-value method.
29. For a right-tailed test and  = 0.01, critical value is t0  2.412 with d.f. = 45. Critical region is values to
the right of 2.412. Since the sample test statistic t = 2.481 is in the critical region, reject H 0 . At the 1%
level, sample data indicate that the average age of Minnesota region coyotes is higher than 1.75 years. The
conclusion is the same as with the P-value method.
30. For a left-tailed test and  = 0.05, critical value is t0  1.676 with d.f. = 50. Critical region is values to
the left of 1.676. Since the sample test statistic t = 1.116 is not in the critical region, fail to reject H 0 . At
the 5% level of significance, the sample data do not indicate that the average fish length is less than 19
inches. The conclusion is the same as with the P-value method.
Section 9.3
1.
2.
The value of p comes from H0. Note that q = 1 – p.
pˆ 
r
n
3.
Yes. The corresponding P value for a one-tailed test is half that of a two-tailed test. Thus the P value for the
one-tailed test is also less than 0.01.
4.
Answers vary. First, we don’t know if the study is based on population data or sample data. If the study is
based on population data, we do not need to conduct a hypothesis test. However, assuming that the study
was based on sample data, testing H0: p = 0.15 versus H1: p > 0.15 seems appropriate. Without a specific
source for the study, we do not know how reliable the information in the sample is. Also, we are not given
information about the sample size. Level of significance could be one of the common values, α = 0.01 or α
= 0.05. Finally, if the conclusions are based on sample data, we cannot conclude that the conclusions are
absolutely true.
5.
(i) (a)   0.01
H 0 : p  0.301
H1: p  0.301
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is
sufficiently large, which it is here, because np = 215(0.301)  647 and nq = 215(0.699)  150.3 are
both > 5.
r
46
pˆ  
 0.214
n 215
pˆ  p 0.214  0.301
z

 2.78
pq
n
0.301(0.699)
215
(c) P value = P(z < 2.78) = 0.0027
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Part IV: Complete Solutions, Chapter 9
503
(d) Since a P value of 0.0027  0.01, we reject H 0 . Yes, the data are statistically significant.
(e) At the 1% level of significance, the sample data indicate that the population proportion of numbers
with a leading 1 in the revenue file is less than 0.301 predicted by Benford’s law.
(ii) Yes. The revenue data file seems to include more numbers with higher first nonzero digits than
Benford’s law predicts.
(iii) We have not proved H 0 to be false. However, because our sample data lead us to reject H 0 and
conclude that there are too few numbers with a leading digit 1, more investigation is merited.
6.
(i) (a)   0.01
H 0 : p  0.301
H1: p  0.301
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is
sufficiently large, which it is, because np = 228(0.301)  68.6 and nq = 228(0.699)  159.4 are
both > 5.
r
92
pˆ  
 0.404
n 228
pˆ  p 0.404  0.301
z

 3.39
pq
n
0.301(0.699)
228
(c) P value = P(z > 3.39) = 0.0003
Standard Normal Distribution
-3.50
-3.00
-2.00
-1.00
0.00
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1.00
2.00
3.00 3.39
Part IV: Complete Solutions, Chapter 9
504
(d) Since a P value of 0.0003  0.01, we reject H 0 . Yes, the data are statistically significant.
(e) At the 1% level of significance, the sample data indicate that the proportion of numbers in the
revenue file with a leading digit 1 exceed the 0.301 predicted by Benford’s law.
(ii) Yes. There seem to be too many entries with a leading digit 1.
(iii) We have not proved H 0 to be false. However, because our data led us to reject H 0 and conclude that
there are “too many” numbers with a leading digit 1, more investigation is merited.
7.
(a)   0.01
H 0 : p  0.70
H1: p  0.70
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 32(0.7) = 22.4 and nq = 32(0.3) = 9.6, and both are greater than 5.
r 24
pˆ  
 0.75
n 32
pˆ  p 0.75  0.70
z

 0.62
pq
n
0.70(0.30)
32
(c) P value = 2P(z > 0.62) = 2(0.2676) = 0.5352
(d) Since a P value of 0.5352 > 0.01, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 1% level of significance, we cannot say that the population proportion of arrests of males aged
15 to 34 in Rock Springs is different than 70%.
8.
(a)   0.05
H 0 : p  0.67
H1: p  0.67
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 38(0.67) = 25.46 and nq = 38(0.33) = 12.54, and both are greater
than 5.
r 21
pˆ  
 0.5526
n 38
pˆ  p 0.5526  0.67
z

 1.54
pq
n
0.67(0.33)
38
(c) P value = P(z < 1.54) = 0.0618
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505
(d) Since a P value of 0.0618 > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 5% level of significance, there is insufficient evidence to say that the proportion of women
athletes who graduate is less than 67%.
9.
(a)   0.01
H 0 : p  0.77
H1: p  0.77
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 27(0.77) = 20.79 and nq = 27(0.23) = 6.21, and both are greater
than 5.
r 15
pˆ  
 0.5556
n 27
pˆ  p 0.5556  0.77
z

 2.65
pq
n
0.77(0.23)
27
(c) P value = P(z < 2.65) = 0.0004
(d) Since a P value of 0.0004  0.01, we reject H 0 . Yes, the data are statistically significant.
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506
(e) At the 1% level of significance, the data show that the population proportion of driver fatalities related
to alcohol is less than 77% in Kit Carson County.
10. (a)   0.05
H 0 : p  0.24
H1: p  0.24
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 56(0.24) = 13.44 and nq = 56(0.76) = 42.56, and both are greater
than 5.
r 12
pˆ  
 0.2143
n 56
pˆ  p 0.2143  0.24
z

 0.45
pq
n
0.24(0.76)
56
(c) P value = 2P(z < 0.45) = 2(0.3264) = 0.6528
(d) Since a P value of 0.6528 > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 5% level of significance, the data do not indicate that the proportion of college students favoring
the color blue is different from 0.24.
11. (a)   0.01
H 0 : p  0.50
H1: p  0.50
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 34(0.50) = 17 and nq = 17, and both are greater than 5.
r 10
pˆ  
 0.2941
n 34
pˆ  p 0.2941  0.50
z

 2.40
pq
n
0.5(0.5)
34
(c) P value = P(z < 2.40) = 0.0082
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507
(d) Since a P value of 0.0082  0.01, we reject H 0 . Yes, the data are statistically significant.
(e) At the 1% level of significance, the data indicate that the population proportion of female wolves is
now less than 50% in the region.
12. (a)   0.05
H 0 : p  0.75
H1: p  0.75
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 83(0.75) = 62.25 and nq = 83(0.25) = 20.75, and both are greater
than 5.
r 64
pˆ  
 0.7711
n 83
pˆ  p 0.7711  0.75
z

 0.44
pq
n
0.75(0.25)
83
(c) P value = 2P(z > 0.44) = 2(0.3300) = 0.6600
(d) Since a P value of 0.6600 > 0.05, we fail to reject H 0 . No, the data are not statistically different.
(e) At the 5% level of significance, there is insufficient evidence to indicate that the population proportion
of guests who catch pike of 20 pounds is different from 75%.
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13. (a)   0.01
H 0 : p  0.261
H1: p  0.261
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 317(0.261) = 82.737 and nq = 317(0.739) = 234.263, and both are
greater than 5.
r
61
pˆ  
 0.1924
n 317
pˆ  p 0.1924  0.261
z

 2.78
pq
n
0.261(0.739)
317
(c) P value = 2P(z < 2.78) = 2(0.0027) = 0.0054
(d) Since a P value of 0.0054  0.01, we reject H 0 . Yes, the data are statistically significant.
(e) At the 1% level of significance, the sample data indicate that the population proportion of the fivesyllable sequence is different from the text of Plato’s Republic.
14. (a)   0.01
H 0 : p  0.214
H1: p  0.214
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 493(0.214) = 105.502 and nq = 493(0.786) = 387.498, and both
are greater than 5.
r 136
pˆ  
 0.2759
n 493
pˆ  p 0.2759  0.214
z

 3.35
pq
n
0.214(0.786)
493
(c) P value = P(z > 3.35) = 0.0004
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509
Standard Normal Distribution
-4.00
-3.00
-2.00
-1.00
0.00
1.00
2.00
3.35
4.00
(d) Since a P value of 0.0004  0.01, we reject H 0 . Yes, the data are statistically significant.
(e) At the 1% level of significance, the sample data indicate that the population proportion of the fivesyllable sequence is higher than that found in Plato’s Symposium.
15. (a)   0.01
H 0 : p  0.47
H1: p  0.47
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 1006(0.47) = 472.82 and nq = 1006(0.53) = 533.18, and both are
greater than 5.
r
490
pˆ  
 0.4871
n 1006
pˆ  p 0.4871  0.47
z

 1.09
pq
n
0.47(0.53)
1006
(c) P value = P(z > 1.09) = 0.1379
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(d) Since a P of 0.1379 > 0.01, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 1% level of significance, there is insufficient evidence to support the claim that the population
proportion of customers loyal to Chevrolet is more than 47%.
16. (a)   0.05
H 0 : p  0.80
H1: p  0.80
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 115(0.8) = 92 and nq = 115(0.2) = 23, and both are greater than 5.
r 88
pˆ  
 0.7652
n 115
pˆ  p 0.7652  0.80
z

 0.93
pq
n
0.80(0.20)
115
(c) P value = P(z < 0.93) = 0.1762
(d) Since a P value of 0.1762 > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 5% level of significance, there is insufficient evidence that the population proportion of prices
ending with digits 9 or 5 is less than 80%.
17. (a)   0.05
H 0 : p  0.092
H1: p  0.092
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 196(0.092) = 18.032 and nq = 196(0.908) = 177.968, and both are
greater than 5.
r 29
pˆ  
 0.1480
n 196
pˆ  p 0.1480  0.092
z

 2.71
pq
n
0.092(0.908)
196
(c) P value = P(z > 2.71) = 0.0034
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(d) Since a P value of 0.0034  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the data indicate that the population proportion of students with
hypertension during final exams week is higher than 9.2%.
18. (a)   0.01
H 0 : p  0.12
H1: p  0.12
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 209(0.12) = 25.08 and nq = 209(0.88) = 183.92, and both are
greater than 5.
r
16
pˆ  
 0.0766
n 209
pˆ  p 0.0766  0.12
z

 1.93
pq
n
0.12(0.88)
209
(c) P value = P(z < 1.93) = 0.0268
(d) Since a P value of 0.0268 > 0.01, we fail to reject H 0 . No, the data are not statistically significant.
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512
(e) At the 1% level of significance, the data are insufficient to conclude that the population proportion of
patients having headaches is less than 0.12.
19. (a)   0.01
H 0 : p  0.82
H1: p  0.82
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 73(0.82) = 59.86 and nq = 73(0.18) = 13.14, and both are greater
than 5.
r 56
pˆ  
 0.7671
n 73
pˆ  p 0.7671  0.82
z

 1.18
pq
n
0.82(0.18)
73
(c) P value = 2P(z  1.18) = 2(0.1190) = 0.2380
(d) Since a P value of 0.2380 > 0.01, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 1% level of significance, the evidence is insufficient to indicate that the population proportion
of extroverts among college student government leaders is different from 82%.
20. For a two-tailed test and  = 0.01, critical values are  z0  2.58. The critical regions are values greater
than 2.58 and values less than 2.58. The sample test statistic z = 0.62 is not in the critical region, so we do
not reject H 0 . Result is consistent with the P-value conclusion.
21. For a left-tailed test and  = 0.01, critical value is z0  2.33. The critical region consists of values less
than 2.33. The sample test statistic z = 2.65 is in the critical region, so we reject H 0 . Result is consistent
with the P-value conclusion.
22. For a two-tailed test and  = 0.01, critical value is z0  2.33. The critical region consists of values greater
than 2.33. The sample test statistic z = 1.09 is not in the critical region, so we fail to reject H 0 . Result is
consistent with the P-value conclusion.
Section 9.4
1.
Paired data are dependent.
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Part IV: Complete Solutions, Chapter 9
513
2.
Take the difference of corresponding paired data values. The sample test statistics is d , which is the mean
of the differences.
3.
H0: d  0 . We test that the mean difference is 0.
4.
Here, n is the number of data pairs.
5.
Here, d.f. = n – 1.
6.
(a) For a right-tailed test that the before score is higher, use d = B – A.
(b) For a left-tailed test that the before score is higher, use d = A – B.
7.
(a)   0.05
H 0 : d  0
H1: d  0
Since  is in H1 , a two-tailed test is used.
(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,
symmetric distribution.
Pair
d=BA
1
3
2
2
3
5
4
4
5
10
6
15
7
6
8
7
d  2.25, sd  7.78
t
d  d 2.25  0

 0.818
sd
n
7.78
8
(c) d.f. = n  1 = 8  1 = 7
From Table 6 in Appendix II, 0.818 falls between entries 0.711 and 1.254. Using two-tailed areas, find
that
0.250 < P value < 0.500. Using a TI-84, P value ≈ 0.4402.
(d) Since the P-value interval is > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
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(e) At the 5% level of significance, the evidence is insufficient to claim a difference in population mean
percentage increases for corporate revenue and CEO salary.
8.
(a)   0.01
H 0 : d  0
H1: d  0
Since  is in H1 , a two-tailed test is used.
(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,
symmetric distribution.
Pair
d=BA
1
0.1
2
0.4
3
0.4
4
1.0
5
0.6
6
0.6
7
0.5
d  0.37, sd  0.47
t
d  d
sd
n

0.37  0
0.47
7
 2.08
(c) d.f. = n  1 = 7  1 = 6
From Table 6 in Appendix II, 2.08 falls between entries 1.943 and 2.447. Using two-tailed areas, find
that
0.050 < P value < 0.100. Using a TI-84, the P value ≈ 0.0823.
(d) Since the P-value interval is > 0.01, we do not reject H 0 . No, the data are not statistically significant.
(e) At the 1% level of significance, the evidence is insufficient to claim that there is a difference in
population mean hours per fish between boat fishing and shore fishing.
9.
(a)   0.01
H 0 : d  0
H1: d  0
Since > is in H1 , a right-tailed test is used.
(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,
symmetric distribution.
Pair
d = Jan  April
1
35
2
9
3
26
4
24
5
17
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515
d  12.6, sd  22.66
t
d  d 12.6  0

 1.243
sd
22.66
5
n
(c) d.f. = n  1 = 5  1 = 4
From Table 6 in Appendix II, 1.243 falls between entries 0.741 and 1.344. Using one-tailed areas, find
that
0.125 < P value < 0.250. Using a TI-84, the P value ≈ 0.1408.
(d) Since the P-value interval is > 0.01, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 1% level of significance, the evidence is insufficient to claim average peak wind gusts are
higher in January.
10. (a)   0.05
H 0 : d  0
H1: d  0
Since > is in H1 , a right-tailed test is used.
(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,
symmetric distribution.
Pair
d=BA
1
1.2
2
1.2
3
2.9
4
1.7
5
13.4
6
2.8
7
5.3
8
7.9
9
6.7
10
4.3
d  4.50, sd  4.1226
t
d  d
sd
n

4.50  0
4.1226
10
 3.452
(c) d.f. = n  1 = 10  1 = 9
From Table 6 in Appendix II, 3.452 falls between entries 3.250 and 4.781. Using two-tailed areas, find
that
0.0005 < P value < 0.005. Using a TI-84, P value ≈ 0.0036.
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Part IV: Complete Solutions, Chapter 9
516
Student's t Distribution, d.f. = 9
-4.000
-3.000
-2.000
-1.000
0.000
1.000
2.000
3.452 4.000
(d) Since the P-value interval is  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence is sufficient to claim that the January mean population of
deer has dropped. Note that this test does not determine the cause of the drop. Development around the
highway, disease, etc. may have affected the deer population.
11. (a)   0.05
H 0 : d  0
H1: d  0
Since > is in H1 , a right-tailed test is used.
(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,
symmetric distribution.
Pair
d = winter  summer
1
19
2
4
3
17
4
7
5
9
6
0
7
9
8
10
d  6.125, sd  9.83
t
d  d 6.125  0

 1.76
sd
n
9.83
8
(c) d.f. = n  1 = 8  1 = 7
From Table 6 in Appendix II, 1.76 falls to the between entries 1.617 and 1.895. Using one-tailed areas,
find that
0.05 < P value < 0.075. Using a TI-84, P value ≈ 0.0607.
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Part IV: Complete Solutions, Chapter 9
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(d) Since the P-value interval is > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 5% level of significance, the evidence is insufficient to indicate that the population average
percentage of male wolves in winter is higher.
12. (a)   0.01
H 0 : d  0
H1: d  0
Since  is in H1 , a two-tailed test is used.
(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,
symmetric distribution.
Pair
d=AB
1
2.9
2
1.1
3
2.1
10
1.3
4
2.1
11
4.0
5
1.4
12
5.6
6
1.8
13
7.6
7
3.3
8
5.1
14
0.9
9
1.6
15
4.1
16
6.5
d  1.10, sd  3.745
t
d  d 1.10  0

 1.175
sd
n
3.745
16
(c) d.f. = n  1 = 16  1 = 15
From Table 6 in Appendix II, 1.175 falls between entries 0.691 and 1.197. Use two-tailed areas to find
that
0.250 < P value < 0.500. Using a TI-84, P value ≈ 0.2584.
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(d) Since the P-value interval is > 0.01, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 1% level of significance, the evidence is insufficient to claim that the population average birth
and death rates are different in this region.
13. (a)   0.05
H 0 : d  0
H1: d  0
Since > is in H1 , a right-tailed test is used.
(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,
symmetric distribution.
Pair
d = houses  hogans
1
5
2
2
3
22
4
23
5
4
6
19
7
33
8
32
d  6, sd  21.5
t
d  d 6  0

 0.789
sd
n
21.5
8
(c) d.f. = n  1 = 8  1 = 7
From Table 6 in Appendix II, 0.789 falls between entries 0.711 and 1.254. Use one-tailed areas to find
that
0.125 < P value < 0.250. Using a TI-84, P value ≈ 0.2282.
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Part IV: Complete Solutions, Chapter 9
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(d) Since the P-value interval is > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
(e)
At the 5% level of significance, the evidence is insufficient to show that the mean
number of inhabited houses is greater than that of hogans.
14. (a)   0.05
H 0 : d  0
H1: d  0
Since > is in H1 , a right-tailed test is used.
(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,
symmetric distribution.
Pair
d = flaked  nonflaked
1
4
2
1
3
9
4
2
5
3
6
6
7
21
8
13
d  6.125, sd  8.079
t
d  d 6.125  0

 2.144
sd
n
8.079
8
(c) d.f. = n  1 = 8  1 = 7
From Table 6 in Appendix II, 2.144 falls between entries 1.895 and 2.365. Use one-tailed areas to find
that
0.025 < P value < 0.05. Using a TI-84, P value ≈ 0.0346.
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Part IV: Complete Solutions, Chapter 9
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(d) Since the P-value interval is  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence is sufficient to indicate that the population average
number of flaked stone tools is higher.
15. (i) Use a calculator to verify. Rounded values are used in part (ii).
(ii) (a)   0.05
H0:   0
H1:   0
Since > is in H1 , a right-tailed test is used.
(b) Use the Student’s t distribution. The sample size is greater than 30.
d  2.472, sd  12.124
t
d  d 2.472  0

 1.223
sd
n
12.124
36
(c) d.f. = n  1 = 36  1 = 35
From Table 6 in Appendix II, 1.223 falls between entries 1.170 and 1.306. Use one-tailed
areas to find that
0.100 < P value < 0.125. Using a TI-84, P value ≈ 0.1147.
(d) Since the P-value interval > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 5% level of significance, the evidence is insufficient to claim that the population mean cost
of living index for housing is higher than that for groceries.
16. (i) Use a calculator to verify. Rounded values are used in part (ii).
(ii) (a)   0.05
H0:   0
H1:   0
Since < is in H1 , a left-tailed test is used.
(b) Use the Student’s t distribution. The sample size is greater than 30.
d  5.7391, sd  15.910
t
d  d 5.7391  0

 2.447
sd
n
15.910
46
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Part IV: Complete Solutions, Chapter 9
521
(c) d.f. = 45
From Table 6 in Appendix II, 2.447 falls between entries 2.412 and 2.690. Use one-tailed
areas to find that
0.005 < P value < 0.010. Using a TI-84, P value ≈ 0.0092.
(d) Since the P-value interval  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence is sufficient to claim that the population mean cost of
living index for utilities is less than that for transportation in this region.
17. (a)   0.05
H 0 : d  0
H1: d  0
Since > is in H1 , a right-tailed test is used.
(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,
symmetric distribution.
Pair
d=BA
1
7
2
2
3
9
4
0
5
6
6
1
7
3
8
3
9
3
d  2.0, sd  4.5
t
d  d
sd
n

2.0  0
4.5
9
 1.33
(c) d.f. = n  1 = 9  1 = 8
From Table 6 in Appendix II, 1.33 falls between entries 1.240 and 1.397. Use one-tailed areas to find
that
0.100 < P value < 0.125. Using a TI-84, P value ≈ 0.1096.
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Part IV: Complete Solutions, Chapter 9
522
(d) Since the P-value interval is > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 5% level of significance, the evidence is insufficient to claim that the population score on the
last round is higher than that on the first.
18. (a)   0.05
H 0 : d  0
H1: d  0
Since > is in H1 , a right-tailed test is used.
(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,
symmetric distribution.
Pair
d=15
1
0.6
2
0.5
3
0.1
4
0.2
5
0.7
6
0.9
d  0.4, sd  0.447
t
d  d
sd
n

0.4  0
0.447
6
 2.192
(c) d.f. = n  1 = 6  1 = 5
From Table 6 in Appendix II, 2.192 falls between entries 2.015 and 2.571. Use one-tailed areas to find
that
0.025 < P value < 0.050. Using a TI-84, P value ≈ 0.0400.
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(d) Since the P-value interval is  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence is sufficient to claim that the population mean time for
rats receiving larger rewards to run the maze is less.
19. (a)   0.05
H 0 : d  0
H1: d  0
Since > is in H1 , a right-tailed test is used.
(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,
symmetric distribution.
Pair
d = time 1  time 5
1
1.4
2
1.7
3
0.8
4
1.5
5
0.5
6
0.1
7
1.7
8
1.3
d  0.775, sd  1.0539
t
d  d 0.775  0

 2.080
sd
n
1.0539
8
(c) d.f. = n  1 = 8  1 = 7
From Table 6 in Appendix II, 2.080 falls between entries 1.895 and 2.365. Use one-tailed areas to find
0.025 < P value < 0.050. Using a TI-84, P value ≈ 0.0380.
(d) Since the P-value interval is  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence is sufficient to claim that the population mean time for
rats receiving larger rewards to climb the ladder is less.
20. (a) d  2.25; sd  7.778 ; t0.95 = 2.365 with d.f. = 7; E  2.365; 95% confidence interval for d from –4.25
to 8.75. We are 95% confident that the difference in population mean percentage increase between
company revenue and CEO salaries is between –4.25% and 8.75%.
(b)  = 0.05; c = 1 – = 0.95; H0: d = 0; H1: d ≠ 0; Since d = 0 from the null hypothesis is in the 95%
confidence interval, do not reject H0 at the 5% level of significance. The data do not indicate a
difference in population mean percentage increases between company revenue and CEO salaries. This
result is consistent with the conclusion reached using the P-value method of testing for  = 0.05 in
Problem 7.
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Part IV: Complete Solutions, Chapter 9
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21. For a two-tailed test with  = 0.05 and d.f. = 7, critical values are t0  2.365. The sample test statistic
t = 0.818 is between 2.365 and 2.365, so we do not reject H 0 . This conclusion is the same as that reached
by the P-value method.
22. For a right-tailed test with  = 0.01 and d.f. = 4, the critical value is t0  3.747. The sample test statistic
t = 1.243 is to the left of t0 , so we do not reject H 0 . This conclusion is the same as that reached by the
P-value method.
Section 9.5
1.
The test statistic is x1  x2 .
2.
Use the Student’s t distribution.
3.
H0: μ1 = μ2 or H0: μ1 – μ2 = 0.
4.
r
r
The test statistic is pˆ1  pˆ 2 , where pˆ1  1 and pˆ 2  2 .
n1
n2
5.
r r
The best estimate is p  1 2 .
n1  n2
6.
(a) Satterthwaite’s gives the larger degrees of freedom. It is more conservative to use a smaller degrees of
freedom because there will be more area in the tails of the Student’s t distribution, resulting in larger P
values.
(b) The P value will be larger. Using a larger P value might mean that the test is not significant, whereas a
smaller P value might result in significance.
7.
H1: μ1 > μ2 or H1: μ1 – μ2 > 0.
8.
H1: μ1 < μ2 or H1: μ1 – μ2 < 0.
9.
(a)   0.01
H 0 : 1  2
H1: 1  2
(b) Use the standard normal distribution. We assume that both population distributions are approximately
normal and that 1 and  2 are known.
x1  x2  2.8  2.1  0.7
z
( x1  x2 )  ( 1  2 )
12
 22

0.7  0
0.52
10
n
2
(c) P value = P(z > 2.57) = 0.0051
n1
2
 2.57
 0.7
10
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(d) Since P value = 0.0051  0.01, we reject H 0 . Yes, the data are statistically significant.
(e) At the 1% level of significance, the evidence is sufficient to indicate that the population mean REM
sleep time for children is more than for adults.
10. (a)   0.01
H 0 : 1  2
H1: 1  2
(b) Use the standard normal distribution. We assume that both population distributions are approximately
normal and that 1 and  2 are known.
x1  x2  43  36  7
z
( x1  x2 )  ( 1  2 )
 22

70
 0.96
212  152

12
14
n1
n2
(c) P value = 2P(z > 0.96) = 2(0.1685) = 0.3370
12
(d) Since P value = 0.3370 > 0.01, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 1% level of significance, the evidence is insufficient to claim that there is a difference in mean
population pollution index for Englewood and Denver.
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11. (a)   0.05
H 0 : 1  2
H1: 1  2
(b) Use the standard normal distribution. We assume that both population distributions are approximately
normal and that 1 and  2 are known.
x1  x2  4.9  4.3  0.6
z
( x1  x2 )  ( 1  2 )
12
 22

0.6  0
1.52
46
2
 2.16
 1.2
51
n
2
(c) P value = 2P(z > 2.16) = 2(0.0154) = 0.0308
n1
(d) Since P value = 0.0308  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence is sufficient to show that there is a difference between
mean response regarding preference for camping or fishing.
12. (a)   0.05
H 0 : 1  2
H1: 1  2
(b) Use the standard normal distribution. We assume that both population distributions are approximately
normal and that 1 and  2 are known.
x1  x2  15.2  19.7  4.5
z
( x1  x2 )  ( 1  2 )
12
n1
 22
n
2

4.5  0
7.22
32
2
 5.2
35
 2.91
(c) P value = P(z < 2.91) = 0.0018
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Part IV: Complete Solutions, Chapter 9
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(d) Since P value = 0.0018  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence is sufficient to show that the population mean percentage
of young adults who attend college is higher.
13. (i) Use a calculator to verify. Use rounded results to compute t.
(ii) (a)   0.01
H 0 : 1  2
H1: 1  2
(b) Use the Student’s t distribution. We assume that both population distributions are approximately
normal and that 1 and  2 are unknown.
x1  x2  3.51  3.87  0.36
t
( x1  x2 )  ( 1  2 )
s12
n1
s22
n
2

0.36  0
0.812
10
 0.94
12
2
 0.965
(c) Since n1  n2 , d . f .  n1  1  10  1  9
In Table 6 in Appendix II, 0.965 falls between entries 0.703 and 1.230. Use one-tailed
areas to find that 0.125 < P value < 0.250. Using a TI-84, d.f. ≈ 19.96, and P value ≈ 0.1731.
(d) Since the P-value interval > 0.01, we fail to reject H 0 . No, the data are not statistically significant.
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(e) At the 1% level of significance, the evidence is insufficient to indicate that violent crime in the
Rocky Mountain region is higher than in New England.
14. (i) Use a calculator to verify. Use rounded results to compute t.
(ii) (a)   0.05
H 0 : 1  2
H1: 1  2
(b) Use the Student’s t distribution. We assume that both population distributions are approximately
normal and that 1 and  2 are unknown.
x1  x2  109.50  99.36  10.14
t
( x1  x2 )  ( 1  2 )
s12
s22
10.14  0

15.412
16
n
n1
2
2
 11.57
14
 2.053
(c) Since n1  n2 , d . f .  n1  1  14  1  13 .
In Table 6 in Appendix II, 2.053 falls between entries 1.771 and 2.160. Use one-tailed areas
find that 0.025 < P value < 0.050. Using a TI-84, d.f. ≈ 27.42, and P value ≈ 0.0249.
to
(d) Since the P-value interval is  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence is sufficient to indicate that the population mean rate of
hay fever is lower for the age group over 50.
15. (a)   0.05
H 0 : 1  2
H1: 1  2
(b) Use the Student’s t distribution. Both sample sizes are large, ni  30, and 1 and  2 are unknown.
x1  x2  344.5  354.2  9.7
t
( x1  x2 )  ( 1  2 )
s12
n1
s22
n
2

9.7  0
49.12
30

50.92
30
 0.751
(c) Since n1  n2 , d.f. = n  1 = 30  1 = 29.
From Table 6 in Appendix II, 0.751 falls between entries 0.683 and 1.174. Use two-tailed areas to find
0.250 < P value < 0.500. Using a TI-84, d.f. ≈ 57.92, and P value ≈ 0.4556.
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Part IV: Complete Solutions, Chapter 9
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(d) Since the P-value interval > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 5% level of significance, the evidence is insufficient to indicate there is a difference between the
control and experimental groups in the mean score on the vocabulary portion of the test.
16. (a)   0.01
H 0 : 1  2
H1: 1  2
(b) Use the Student’s t distribution. Both sample sizes are large, ni  30, and 1 and  2 are unknown.
x1  x2  368.4  349.2  19.2
t
( x1  x2 )  ( 1  2 )
s12
n1
s22
n
2

19.2  0
39.52
30

56.62
30
 1.524
(c) Since n1  n2 , d.f. = n  1 = 30  1 = 29.
From Table 6 in Appendix II, 1.524 falls between entries 1.479 and 1.699. Use one-tailed areas to find
0.050 < P value < 0.075. Using a TI-84, d.f. ≈ 51.83, and P value ≈ 0.0668.
(d) Since the P-value interval > 0.01, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 1% level of significance, the evidence is insufficient to claim that the population mean score for
the experimental group was higher than for the control group.
17. (i) Use a calculator to verify. Use rounded results to compute t.
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Part IV: Complete Solutions, Chapter 9
530
(ii) (a)   0.05
H 0 : 1  2
H1: 1  2
(b) Use the Student’s t distribution. We assume that the population distributions for both are
approximately normal and mound-shaped and that 1 and  2 are unknown.
x1  x2  4.75  3.93  0.82
t
( x1  x2 )  ( 1  2 )
s12
n1

s22
n2

0.82  0
2.822
16
2
 0.869
 2.43
15
(c) Since n2  n1 , d . f .  n2  1  15  1  14.
From Table 6 in Appendix II, 0.869 falls between entries 0.692 and 1.200. Use two-tailed
areas to find 0.250 < P value < 0.500. Using a TI-84, d.f. ≈ 28.81, and P value ≈ 0.1961.
(d) Since the P-value interval is > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 5% level of significance, the evidence is insufficient to indicate that there is a difference in
the mean number of cases of fox rabies between the two regions.
18. (i) Use a calculator to verify. Use rounded results to compute t.
(ii) (a)   0.05
H 0 : 1  2
H1: 1  2
(b) Use the Student’s t distribution. We assume the population distributions for both are
approximately normal and mound-shaped and that 1 and  2 are unknown.
x1  x2  12.53  10.77  1.76
t
( x1  x2 )  ( 1  2 )
s12
n1
s22
n
2

1.76  0
2.392
14

2.402
16
 2.008
(c) Since n1  n2 , d . f .  n1  1  14  1  13.
From Table 6 in Appendix II, 2.008 falls between entries 1.771 and 2.160. Use one-tailed
areas to find 0.025 < P value < 0.050. Using a TI-84, d.f. ≈ 27.50, and P value ≈ 0.0273.
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Part IV: Complete Solutions, Chapter 9
531
(d) Since the P-value interval is  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence is sufficient to indicate that population mean soil
water content is higher in Field A.
19. (i) Use a calculator to verify. Use rounded results to compute t.
(ii) (a)   0.05
H 0 : 1  2
H1: 1  2
(b) Use the Student’s t distribution. We assume the population distributions for both are
approximately normal and mound-shaped and that 1 and  2 are unknown.
x1  x2  4.86  6.5  1.64
t
( x1  x2 )  ( 1  2 )
s12
n1
s22
n
2

1.64  0
3.182
7

2.882
8
 1.041
(c) Since n1  n2 , d . f .  n1  1  7  1  6.
From Table 6 in Appendix II, 1.041 falls between entries 0.718 and 1.273. Use two-tailed
areas to find 0.250 < P value < 0.500. Using a TI-84, d.f. ≈ 12.28, and P value ≈ 0.3179.
Copyright © Houghton Mifflin Company. All rights reserved.
Part IV: Complete Solutions, Chapter 9
532
(d) Since the P-value interval is > 0.05, we fail to reject H 0 . No, the data are not statistically
significant.
(e) At the 5% level of significance, the evidence is insufficient to indicate that the mean time lost
owing to hot tempers is different from that lost owing to technical workers’ attitudes.
20.
(a) d  2.25; sd  7.778; t0.95 = 2.365 with d.f. = 7; E  2.365; 95% confidence interval for d from –
4.25 to 8.75. We are 95% confident that the difference in population mean percentage increase
between company revenue and CEO salaries is between –4.25% and 8.75%.
(b)  = 0.05; c = 1 –  = 0.95; H0: d = 0; H1: d ≠ 0; since d = 0 from the null hypothesis is in the
95% confidence interval, do not reject H0 at the 5% level of significance. The data do not indicate
a difference in population mean percentage increases between company revenue and CEO salaries.
This result is consistent with the conclusion reached using the P-value method of testing for  =
0.05 in Problem 7.
21. (a) d . f . 
 s12 s22 
 n1  n2 


2



2
0.812  0.942
10
12
2
2
1 0.812
1 0.942
 11
9 10
12
 19.96
2
2

 s22 
1
  n2 1  n2 

 
(Note: Some software will truncate this to 19.)
(b) In Problem 13, d . f .  n1  1  10  1  9.
The convention of using the smaller of n1  1 and n2  1 leads to a d.f. that is always less than or equal to
that computed by Satterthwaite’s formula.
1  s1
n1 1  n1
2
(n1  1) s12  (n2  1) s22
22. (a) s 
n1  n2  2
x1  x2
t
 
15(2.822 )  14(2.432 )
 2.639
16  15  2
4.75  3.93
 0.865
1  1
2.639 16
15
d . f .  n1  n2  2  16  15  2  29
H0 : 1  2
H1: 1  2
From Table 6 in Appendix II, 0.865 falls between entries 0.683 and 1.174. Using two-tailed areas,
find 0.250 < P value < 0.500. Since the P-value interval > 0.05, we fail to reject H 0 .
(b) Using unpooled sample standard deviation, t  0.869, d.f. = 14, and fail to reject H 0 . Conclusions are
the same, however; the exact P value using the pooled standard deviation method and larger d.f. will be
slightly smaller than that using unpooled methods.
s
1
n1

1
n2


 
23. (a)   0.05
H 0 : p1  p2
H1: p1  p2
(b) Use the standard normal distribution. The number of trials is sufficiently large because n1 p, n1q , n2 p,
and n2 q are each larger than 5.
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Part IV: Complete Solutions, Chapter 9
533
r1  r2
59  56

 0.2911
n1  n2 220  175
q  1  p  1  0.2911  0.7089
p
r1
r
59
56

 0.268, pˆ 2  2 
 0.32
n1 220
n2 175
pˆ1  pˆ 2  0.268  0.32  0.052
pˆ  pˆ 2
0.052
z 1

 1.13
pq
pq
0.2911(0.7089) 0.2911(0.7089)


n
n
220
175
pˆ1 
1
2
(c) P value = 2P(z < 1.13) = 2(0.1292) = 0.2584
(d) Since the P value = 0.2584 > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 1% level of significance, there is insufficient evidence to conclude that the population
proportion of women favoring more tax dollars for the arts is different from the proportion of men.
24. (a)   0.05
H 0 : p1  p2
H1: p1  p2
(b) Use the standard normal distribution. The number of trials is sufficiently large because n1 p, n1q , n2 p,
and n2 q are each larger than 5.
r1  r2 21  22

 0.2443
n1  n2 93  83
q  1  p  1  0.2443  0.7557
p
r1 21
r
22

 0.226, pˆ 2  2 
 0.265
n1 93
n2 83
pˆ1  pˆ 2  0.226  0.265  0.039
pˆ  pˆ 2
0.039
z 1

 0.61
pq
pq
0.2443(0.7557) 0.2443(0.7557)


n
n
93
83
pˆ1 
1
2
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Part IV: Complete Solutions, Chapter 9
534
(c) P value = P(z < 0.61) = 0.2709
(d) Since the P value = 0.2709 > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 5% level of significance, there is insufficient evidence to conclude that the population
proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of
moderate voters.
25. (a)   0.01
H 0 : p1  p2
H1: p1  p2
(b) Use the standard normal distribution. The number of trials is sufficiently large because n1 p, n1q , n2 p,
and n2 q are each larger than 5.
r1  r2
12  7

 0.0676
n1  n2 153  128
q  1  p  1  0.0676  0.9324
p
r1 12
r
7

 0.0784, pˆ 2  2 
 0.0547
n1 153
n2 128
pˆ1  pˆ 2  0.0784  0.0547  0.0237
pˆ  pˆ 2
0.0237
z 1

 0.79
pq
pq
0.0676(0.9324) 0.0676(0.9324)


n
n
153
128
pˆ1 
1
2
(c) P value = 2P(z > 0.79) = 2(0.2148) = 0.4296
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Part IV: Complete Solutions, Chapter 9
535
(d) Since the P value = 0.4296 > 0.01, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 1% level of significance, there is insufficient evidence to conclude that the population
proportion of high-school dropouts on Oahu is different from that of Sweetwater County.
26. (a)   0.05
H 0 : p1  p2
H1: p1  p2
(b) Use the standard normal distribution. The number of trials is sufficiently large because n1 p, n1q , n2 p,
and n2 q are each larger than 5.
r1  r2
141  125

 0.5278
n1  n2 288  216
q  1  p  1  0.5278  0.4722
p
r1 141
r
125

 0.4896, pˆ 2  2 
 0.5787
n1 288
n2 216
pˆ1  pˆ 2  0.4896  0.5787  0.0891
pˆ  pˆ 2
0.0891
z 1

 1.98
pq
pq
0.5278(0.4722) 0.5278(0.4722)


n
n
288
216
pˆ1 
1
2
(c) P value = P(z < 1.98) = 0.0239
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Part IV: Complete Solutions, Chapter 9
536
(d) Since the P value = 0.0239 < 0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, there is sufficient evidence to conclude that the population proportion
of voter turnout in Colorado is greater than that in California.
27. (a)   0.01
H 0 : p1  p2
H1: p1  p2
(b) Use the standard normal distribution. The number of trials is sufficiently large because n1 p, n1q , n2 p,
and n2 q are each larger than 5.
r1  r2
37  47

 0.42
n1  n2 100  100
q  1  p  1  0.42  0.58
p
r1
r
37
47

 0.37, pˆ 2  2 
 0.47
n1 100
n2 100
pˆ1  pˆ 2  0.37  0.47  0.10
pˆ  pˆ 2
0.10
z 1

 1.43
pq
pq
0.42(0.58) 0.42(0.58)


n
n
100
100
pˆ1 
1
2
(c) P value = P(z < -1.43)  0.0764
(d) Since the P value = 0.0764 > 0.01, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 1% level of significance, there is insufficient evidence to conclude that the population
proportion of adults believing in extraterrestrials who attended college is higher than the proportion
who did not attend college.
28. (a)   0.05
H 0 : p1  p2
H1: p1  p2
(b) Use the standard normal distribution. The number of trials is sufficiently large because n1 p, n1q , n2 p,
and n2 q are each larger than 5.
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Part IV: Complete Solutions, Chapter 9
537
r1  r2 45  36

 0.6694
n1  n2 59  62
q  1  p  1  0.6694  0.3306
p
r1 45
r
36

 0.7627, pˆ 2  2 
 0.5806
n1 59
n2 62
pˆ1  pˆ 2  0.7627  0.5806  0.182
pˆ  pˆ 2
0.182
z 1

 2.13
pq
pq
0.6694(0.3306) 0.6694(0.3306)


n
n
59
62
pˆ1 
1
2
(c) P value = P(z > 2.13) = 0.0166
(d) Since the P value = 0.0166 < 0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, there is sufficient evidence to conclude that the population proportion
of conservative voters who prefer art with fully clothed people is higher.
29. (a)   0.05
H 0 : p1  p2
H1: p1  p2
(b) Use the standard normal distribution. The number of trials is sufficiently large because n1 p, n1q , n2 p,
and n2 q are each larger than 5.
r1  r2
45  71

 0.2189
n1  n2 250  280
q  1  p  1  0.2189  0.7811
p
r1
r
45
71

 0.18, pˆ 2  2 
 0.2536
n1 250
n2 280
pˆ1  pˆ 2  0.18  0.2536  0.074
pˆ  pˆ 2
0.074
z 1

 2.04
pq
pq
0.2189(0.7811) 0.2189(0.7811)


n
n
250
280
pˆ1 
1
2
(c) P value = P(z < 2.04) = 0.0207
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Part IV: Complete Solutions, Chapter 9
538
(d) Since the P value = 0.0207 < 0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, there is sufficient evidence to conclude that the population proportion
of trusting people in Chicago is higher in the older group.
30. H0 : 1  2 ; H1: 1  2 ; for  = 0.01 and a right-tailed test, the critical value z0  2.33; sample test
statistic z = 2.57 is in the critical region, reject H 0 . This result is consistent with the P-value method.
31. H0 : 1  2 ; H1: 1  2 ; for d.f. = 9,  = 0.01 in the one-tail area row, the critical value t0  2.821.
Sample test statistic t = 0.965 is not in the critical region, fail to reject H 0 . This result is consistent with
the P-value method.
32. H 0 : p1  p2 ; H1: p1  p2 ; for  = 0.05 and a two-tailed test, the critical values are  z0  1.96; sample
test statistic z = 1.13 is not in the critical region, fail to reject H 0 . This result is consistent with the Pvalue method.
Chapter 9 Review
1.
Look at the original x distribution. If it is normal or n ≥ 30, and σ is known, use the standard normal
distribution. If the x distribution is mound-shaped or n ≥ 30, and σ is unknown, use the Student’s t
distribution. The degrees of freedom are determined by the application.
2.
A significant test means we reject H0. Statistically significant results are not necessarily practically
significant results.
3.
A larger sample size increases the value of z or t.
4.
A larger value of z or t will have a smaller P value.
5.
(a)   0.05
H 0 :   11.1
H1:   11.1
Since  is in H1 , use a two-tailed test.
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Part IV: Complete Solutions, Chapter 9
539
(b) Use the standard normal distribution. We assume that x has a normal distribution with known standard
deviation  . Note that  is given in the null hypothesis.
z
x 

n

10.8  11.1
0.6
36
 3.00
(Note: 600 miles = 0.6 thousand miles)
(c) P value = 2P(z < 3.00) = 2(0.0013) = 0.0026
(d) Since a P value of 0.0026  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence is sufficient to say that the miles driven per vehicle in
Chicago is different from the national average.
6.
(a)   0.05
H 0 : p  0.35
H1: p  0.35
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 81(0.35) = 28.35 and nq = 81(0.65) = 52.65, and both are greater
than 5.
r 39
pˆ  
 0.4815
n 81
pˆ  p 0.4815  0.35
z

 2.48
pq
n
0.35(0.65)
81
(c) P value = P(z > 2.48) = 0.0066
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Part IV: Complete Solutions, Chapter 9
540
(d) Since a P value of 0.0066  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence indicates that more than 35% of the students have jobs.
7.
(a)   0.01
H 0 :   0.8 A
H1:   0.8 A
(b) Use the Student’s t distribution with d.f. = n  1 = 9  1 = 8. We assume that x has a distribution that is
approximately normal and that  is unknown.
x   1.4  0.8
t

 4.390
s
n
0.41
9
(c) For d.f. = 8, 4.390 falls between entries 3.355 and 5.041. Using one-tailed areas,
0.0005 < P value < 0.005. Using a TI-84, P value ≈ 0.0012.
Student's t Distribution, d.f. = 8
-3.00
-2.00
-1.00
0.00
1.00
2.00
3.00
4.39 5.00
(d) Since the entire P-value interval  0.01, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence is sufficient to say that the Toylot claim of 0.8 A is too
low.
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Part IV: Complete Solutions, Chapter 9
8.
541
(a)   0.01
H 0 : 1  2
H1: 1  2
(b) Use the Student’s t distribution. We assume that both population distributions are approximately
normal and that 1 and  2 are unknown.
x1  x2  9.4  6.9  2.5
t
( x1  x2 )  ( 1  2 )
s12
n1

s22
n2

2.5
2.12
12
2
 2.986
 2.0
12
(c) Since n1  n2 , d.f. = n  1 = 12  1 = 11.
In Table 6 in Appendix II, 2.986 falls between entries 2.718 and 3.106. Use one-tailed areas to find
that 0.005 < P value < 0.010. Using a TI-84, d.f. ≈ 21.95, and P value ≈ 0.0034.
(d) Since the P-value interval is  0.01, we reject H 0 . Yes, the data are statistically significant.
(e) At the 1% level of significance, the evidence shows that the yellow paint has less visibility after
1 year.
9.
(a)   0.01
H 0 : p  0.60
H1: p  0.60
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 90(0.6) = 54 and nq = 90(0.4) = 36, and both are greater than
5.
r 40
pˆ  
 0.4444
n 90
pˆ  p 0.4444  0.60
z

 3.01
pq
n
0.60(0.40)
90
(c) P value = P(z < 3.01) = 0.0013
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Part IV: Complete Solutions, Chapter 9
542
(d) Since a P value of 0.0013  0.01, we reject H 0 . Yes, the data are statistically significant.
(e) At the 1% level of significance, the evidence is sufficient to show the mortality rate has dropped.
10. (a)   0.05
H 0 : 1  2
H1: 1  2
(b) Use the Student’s t distribution. We assume that both population distributions are approximately
normal and that 1 and  2 are unknown.
x1  x2  53  62  9
t
( x1  x2 )  ( 1  2 )
s12
n1

s22
n2

9  0
192
51
2
 2.735
 15
60
(c) Since n1  n2 , d . f .  n1  1  51  1  50.
In Table 6 in Appendix II, 2.735 falls between entries 2.678 and 3.496. Use two-tailed areas to find
that 0.001 < P value < 0.010. Using a TI-84, d.f. ≈ 94.53, and P value ≈ 0.0074.
(d) Since the P-value interval is  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence shows that there is a significant difference in average offschedule times.
Copyright © Houghton Mifflin Company. All rights reserved.
Part IV: Complete Solutions, Chapter 9
543
11. (a)   0.01
H 0 :   40
H1:   40
Since > is in H1 , use a right-tailed test.
(b) Use the standard normal distribution. We assume that x has a normal distribution with known standard
deviation  . Note that  is given in the null hypothesis.
z
x 


43.1  40
9
94
n
 3.34
(c) P value = P(z > 3.34) = 0.0004
Standard Normal Distribution
-4.00
-3.00
-2.00
-1.00
0.00
1.00
2.00
3.34
4.00
(d) Since a P value of 0.0004  0.01, we reject H 0 . Yes, the data are statistically significant.
(e) At the 1% level of significance, the evidence is sufficient to say that the population average number of
matches is larger than 40.
12. (a)   0.05
H 0 : p1  p2
H1: p1  p2
(b) Use the standard normal distribution. The number of trials is sufficiently large because n1 p, n1q , n2 p,
and n2 q are each larger than 5.
r1  r2 12  18

 0.1622
n1  n2 88  97
q  1  p  1  0.1622  0.8378
p
r1 12
r
18

 0.1364, pˆ 2  2 
 0.1856
n1 88
n2 97
pˆ1  pˆ 2  0.1364  0.1856  0.0492
pˆ  pˆ 2
0.0492
z 1

 0.91
pq
pq
0.1622(0.8378) 0.1622(0.8378)
n

n
88
97
pˆ1 
1
2
(c) P value = P(z < 0.91) = 0.1814
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Part IV: Complete Solutions, Chapter 9
544
(d) Since the P value = 0.1814 > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
(e) At the 5% level of significance, there is insufficient evidence to say that the population proportion of
suburban residents subscribing to Sporting News is higher.
13. (a)   0.05
H 0 : 1  2
H1: 1  2
(b) Use the Student’s t distribution. We assume that both population distributions are approximately
normal and that 1 and  2 are unknown.
x1  x2  3.0  2.7  0.3
t
( x1  x2 )  ( 1  2 )
s12
n1

s22
n2

0.3  0
0.82
55
2
 1.808
 0.9
51
(c) Since n2  n1 , d . f .  n2  1  51  1  50.
In Table 6 in Appendix II, 1.808 falls between entries 1.676 and 2.009. Use two-tailed areas to find
that 0.050 < P value < 0.100. Using a TI-84, d.f. ≈ 100.27, and P value ≈ 0.0735.
(d) Since the P-value interval > 0.05, we fail to reject H 0 . No, the data are not statistically significant.
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Part IV: Complete Solutions, Chapter 9
545
(e) At the 5% level of significance, the evidence is insufficient to show that there is a difference in
population mean length of the two types of projectile points.
14. (a)   0.05
H 0 : p  0.36
H1: p  0.36
(b) Use the standard normal distribution. The p̂ distribution is approximately normal when n is sufficiently
large, which it is here, because np = 120(0.36) = 43.2 and nq = 120(0.64) = 76.8, and both are greater
than 5.
r
33
pˆ  
 0.275
n 120
pˆ  p 0.275  0.36
z

 1.94
pq
n
0.36(0.64)
120
(c) P value = P(z < 1.94) = 0.0262
(d) Since a P value of 0.0262  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, the evidence is sufficient to show that the population percentage of
bachelor or higher degree holders in the private sector is less than in the federal civilian sector.
15. (a)   0.05
H 0 :   7 oz
H1:   7 oz
(b) Use the Student’s t distribution with d.f. = n  1 = 8  1 = 7. We assume that x has a distribution that is
approximately normal and that  is unknown.
x   7.3  7
t

 1.697
s
n
0.5
8
(c) For d.f. = 7, 1.697 falls between entries 1.617 and 1.895. Using two-tailed areas,
0.100 < P value < 0.150. Using a TI-84, P value ≈ 0.1335.
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Part IV: Complete Solutions, Chapter 9
546
(d) Since the entire P-value interval > 0.05, we fail to reject H 0 . No, the data are not statistically
significant.
(e) At the 5% level of significance, the evidence is insufficient to show that the population mean amount
of coffee per cup is different from 7 ounces.
16. (a)   0.01
H 0 : d  0
H1: d  0
(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,
symmetric distribution.
Pair
d = exp  cont
A
14
B
14
C
7
D
7
E
12
F
5
d  9.83, sd  3.97
t
d  d 9.83  0

 6.07
sd
n
3.97
6
(c) d.f. = n  1 = 6  1 = 5
From Table 6 in Appendix II, 6.07 falls between entries 4.032 and 6.869. Using one-tailed areas, find
0.0005 < P value < 0.005. Using a TI-84, P value ≈ 0.0009.
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Part IV: Complete Solutions, Chapter 9
547
Student's t Distribution, d.f. = 5
-3.000 -2.000 -1.000
0.000
1.000
2.000
3.000
4.000
5.000
6.066
7.000
(d) Since the P-value interval is  0.01, we reject H 0 . Yes, the data are statistically significant.
(e) At the 1% level of significance, there is sufficient evidence to indicate that the program of the
experimental group promoted creative problem solving.
17. (a)   0.05
H 0 : d  0
H1: d  0
(b) Use the Student’s t distribution. Assume that d has a normal distribution or has a mound-shaped,
symmetric distribution.
Pair
d = before  after
1
6.4
2
7.2
3
8.6
4
3.8
5
1.3
d  4.94, sd  3.90
t
d  d
sd
n

4.94  0
3.90
5
 2.832
(c) d.f. = n  1 = 5  1 = 4
From Table 6 in Appendix II, 2.832 falls between entries 2.776 and 3.747. Use one-tailed areas to find
that 0.010 < P value < 0.025. Using a TI-84, P value ≈ 0.0236.
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Part IV: Complete Solutions, Chapter 9
548
(d) Since the P-value interval is  0.05, we reject H 0 . Yes, the data are statistically significant.
(e) At the 5% level of significance, there is insufficient evidence to claim that the injection system lasts
less than an average of 48 months.
18. (i) Use a calculator to verify. The rounded values are used in part (ii).
(ii) (a)   0.05
H 0 :   48 months
H1:   48 months
(b) Use the Student’s t distribution with d.f. = n  1 = 10  1 = 9. We assume that x has a distribution
that is approximately normal and that  is unknown.
x   44.2  48
t

 1.396
s
n
8.61
10
(c) For d.f. = 9, 1.396 falls between entries 1.383 and 1.574. Use one-tailed areas to find that
0.075 < P value < 0.100. Using a TI-84, P value ≈ 0.0981.
(d) Since the entire P-value interval > 0.05, we fail to reject H 0 . No, the data are not statistically
significant.
(e) At the 5% level of significance, there is insufficient evidence to claim that the injection system lasts
less than an average of 48 months.
Cumulative Review Chapters 7–9
1.
(a) Answers vary. A population is the entire group of interest. Random sampling is accomplished through
the assigning of numbers to members of the population and then by using a random-number table or a
computer package to draw the simple random sample.
(b) A sample statistic is a numerical value calculated from the sample. A sampling distribution is the
probability distribution associated with a sample statistics such as the sample mean.
(c) See Chapter 7.
(d) Answers vary.
2.
(a) Since the sample size is large, the central limit theorem describes the distribution of x .
(b)
P( x  6820)  P( z  2.75)  0.0030
z
6820  7500
 2.75
1750
50
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Part IV: Complete Solutions, Chapter 9
549
(c) The probability that the 50 workers’ average white blood cell count is that low or lower, if they were
healthy adults, is extremely low. It would be reasonable to gather additional facts.
3.
We are given p = 0.45 and n = 32. Both np > 5 and nq > 5.
(a)
18
0.5
pˆ 
 0.5625 Continuity correction 
 0.0156
32
32
P( pˆ  0.5625)  P(0.5625  pˆ )  P(0.5625  0.0156  x)
P(0.5469  x)  P(1.10  z )  0.1357
0.5469  0.45
z
 1.10
0.45  0.55
32
(b)
10
0.5
 0.3125 Continuity correction 
 0.0156
32
32
P( pˆ  0.3125)  P( pˆ  0.3125  0.0156)
pˆ 
P( x  0.3281)  P( z  1.39)  0.0823
0.3281  0.45
z
 1.39
0.45  0.55
32
(c) Yes. Here, np = 32 × 0.45 = 14.4 and nq = 32 × 0.55 = 17.6, both greater than 5.
The expected value of pˆ is p  0.45 .
The standard deviation of pˆ is  pˆ 
4.
pq

n
0.45  0.55
 0.088 .
32
(a) H0: μ = 2.0
H1: μ > 2.0
α = 0.01
Since σ is known and x is normal, we will use the standard normal distribution.
2.56  2.0
z
 2.53
0.7
10
The P value will be P(z > 2.53) = 0.0057.
Since the P value = 0.0057 < 0.01, we reject H0. Yes, the data are statistically significant.
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Part IV: Complete Solutions, Chapter 9
550
At the 1% level of significance, the evidence is sufficient to say that the population mean discharge
level of lead is higher.
(b)
x  z
n
   x  z
2.56  1.96  0.7
10
n
   2.56  1.96  0.7
10
2.13    2.99
2
2
z
 1.96  0.7 
(c) n   c   
  47.06, so use n = 48 water samples.
E
 0.2 


5.
(a) Use a calculator to confirm that x  10.49 and s ≈ 1.36.
(b) H0: μ = 10%
H1: μ > 10%
α = 0.05
Since x is approximately normal and σ is unknown, we will use a Student’s t distribution.
d.f. = 12 – 1 = 11
10.49  10
t
 1.248
1.36
12
Using a TI-84, the P value will be P(t > 1.248) = 0.119.
Using Table 6 in Appendix II, 0.100 < P value < 0.125.
Since the P-value interval > 0.05, we fail to reject H0. The data are not statistically significant.
At the 5% level of significance, the evidence does not indicate that the patient is asymptotic.
(c) For a 99% interval, tc = 3.106.
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Part IV: Complete Solutions, Chapter 9
x t s
n
   x t s
10.49  3.106  1.36
12
551
n
   10.49  3.106  1.36
12
9.27%    11.71%
6.
(a) Check conditions: np  68  0.10  6.8, nq  68  0.90  61.2. Since both estimates are reasonably
greater than 5, the conditions are satisfied.
H0: p = 0.10
H1: p ≠ 0.10
α = 0.05
Since the conditions are satisfied, p̂ follows a normal distribution.
z
0.147  0.10
 1.29
0.10  0.90
68
The P value will be 2 × P(z > 1.29) = 0.197.
Since the P value > 0.05, we fail to reject H0. No, the data are not statistically significant.
At the 5% level of significance, the data do not indicate any difference from the national average for
the population proportion of crime victims.
(b) For a 95% confidence interval, zc = 1.96.
ˆˆ
ˆˆ
pq
pq
pˆ  z 
 p  pˆ  z 
n
n
0.147  1.96 
0.147  0.853
0.147  0.853
 p  0.147  1.96 
68
68
0.063  p  0.231
2
 1.96 
(c) n  
  0.147  0.853  192.68, so use n = 193 students.
 0.05 
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Part IV: Complete Solutions, Chapter 9
552
7.
H0: μd = 0
H1: μd ≠ 0
α = 0.05
Pair
d=A–B
1
–0.001
2
–0.028
3
–0.002
4
0.016
5
–0.01
6
0.001
7
–0.003
Here, d  0.0039 , sd = 0.013, and we will use the Student’s t distribution with d.f. = 7 – 1 = 6.
0.0039  0
t
 0.771
0.013
7
Using a TI-84, the P value will be 2 × P(t < –0.771) = 0.470.
Using Table 6 in Appendix II, 0.250 < P value < 0.500.
Since the P-value interval > 0.05, we fail to reject H0. No, the data are not statistically significant.
At the 5% level of significance, the evidence does not show a difference in phosphorous reduction between
the two methods.
8.
(a) H0: μ1 = μ2
H 1: μ 1 ≠ μ 2
α = 0.05
We will use the Student’s t distribution with d.f. = 16 – 1 = 15.
 74.8  70.1   0  0 
t
 1.952
 5.22   8.62 



 16   18 

 

Using the TI-84, the P value will be 2 × P(t > 1.952) = 0.070.
Using Table 6 in Appendix II, 0.050 < P value < 0.100.
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Part IV: Complete Solutions, Chapter 9
553
Since the P-value interval > 0.05, we fail to reject H0. No, the data are not statistically significant.
At the 5% level of significance, the evidence does not show any difference in the population mean
proportion of on-time arrivals in summer versus winter.
(b)
 x1  x2   t 
s12 s2 2
s2 s 2

 1   2   x1  x2   t  1  2
n1
n2
n1
n2
 74.8  70.1  2.131
5.22 8.62
5.22 8.62

 1   2   74.8  70.1  2.131

16
18
16
18
0.43%  1   2  9.83%
(c) We assume that x1 and x2 are approximately normal or at least mound-shaped and symmetric.
9.
(a) H0: p1 = p2
H1: p1 > p2
α = 0.05
We check that n1 pˆ1  23, n2 pˆ 2  19, n1qˆ1  72, and n2 qˆ2  73, and all are greater than 5. It thus seems
reasonable to use the normal distribution.
23  19
p
 0.2246
95  92
 0.242  0.207    0  0 
z
 0.58
0.2246  0.7754 0.2246  0.7754

95
92
The P value will be P(z > 0.58) = 0.2810.
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Part IV: Complete Solutions, Chapter 9
554
Since P value > 0.05, we fail to reject H0. No, the data are not statistically significant.
At the 5% level of significance, the evidence does not indicate that the population proportion of single
men is greater than the population proportion of single women.
(b)
 pˆ1  pˆ 2   z 
pq pq

 p1  p2   pˆ1  pˆ 2   z 
n1 n2
pq pq

n1 n2
 0.242  0.207   1.645 
0.2246  0.7754 0.2246  0.7754

 p1  p2 
95
92
 0.242  0.207   1.645 
0.2246  0.7754 0.2246  0.7754

95
92
0.065  p1  p2  0.139
10. (a) Essay. Refer to Chapters 7, 8, and 9.
(b) Answers will vary.
11. Answers will vary. Salient points include that the hypotheses of interest are
H0: μ1 = μ2
H1: μ1 < μ2, with μ1 representing the average time to recovery under the new method
At the 1% level of significance, the data are statistically significant, indicating that the new method leads to
shorter recovery times, on average.
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